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Let $X$ be a smooth complete complex (algebraic) 3-fold, $D$ an effective divisor on $X$, and $C$ a smooth integral curve contained in the support of $D$. Let $X'$ be the blowup of $X$ along $C$, and denote by $D'$ the strict transform of $D$, $E$ the exceptional divisor, and $C'=D' \cap E$ the (set-theoretic) intersection.

Then is it true that $C \cong C'$, $D' \cong D$, and the normal bundles $N_{C/D} \cong N_{C'/D'}$ under the identification of $C \cong C'$ ? Are the degrees of the normal bundles the same?

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I don't think $D' \simeq D$. If $D$ has two components $D_1,D_2$ with transverse intersection $C=D_1 \cap D_2$, then the strict transform under the blow up is disconnected. – J.C. Ottem May 20 '11 at 13:45
J.C. Ottem is correct; on the other hand I think $D^\prime \simeq D$ should be true under the assumption that $D$ is smooth along $C$. – user5117 May 20 '11 at 13:53
Yes, I see now. Thank you for your comments. – Parsa May 21 '11 at 16:21
up vote 6 down vote accepted

Let $X=\mathbb P^3$, $D$ a quadric cone and $C$ a line on $D$ that goes through the vertex of $D$. Then $D'$ is isomorphic to $D$ blown up at its vertex, in particular it is smooth, $E\simeq C\times \mathbb P^1$ and $C'$ is isomorphic to the union of one member of each of the rulings on $E$, that is $C\cup \mathbb P^1$ intersecting in a single point transversally.

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In general $D' \not\cong D$. In fact, $D'$ is the blow-up of $D$ in (the sheaf of ideals of) $C$ and $C'$ is the exceptional divisor of this blowup. So, if $C$ is a Cartier divisor on $D$ (locally given by 1 equation) then $D' \cong D$, $C' \cong C$. This is true if $D$ has multiplicity 1 along $C$.

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Aha, while Sandor's example is excellent and answers the question, this is really answers the question I meant to ask. Thank you. – Parsa May 21 '11 at 16:20

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